Limit of ln(sinx) & Limit of [ln(1+x^2)-ln(1+x)]

  • Thread starter QuarkCharmer
  • Start date
  • Tags
    Limit
In summary, the limit of ln(sin(x)) as x approaches 0+ tends towards -infinity due to the properties of both sin(x) and ln(x). For the limit of [ln(1+x^2)-ln(1+x)] as x approaches infinity, we can use the squeeze theorem and the fact that ln(x) approaches infinity as x approaches infinity to show that it also tends towards infinity.
  • #1
QuarkCharmer
1,051
3

Homework Statement


[itex]lim_{x \to 0+} ln(sin(x))[/itex]

[itex]lim_{x \to \infty} [ln(1+x^2)-ln(1+x)][/itex]

Homework Equations

The Attempt at a Solution


I'm really not sure how to take this limit at all?

I know (from using a table) that it tends towards -infinity, but I am not sure how to go about taking it manually?

I am thinking that because as x approaches 0, sin(x) approaches 0, you can treat sin(x) like x. Then as x approaches 0, ln(x) approaches -infinity.

For the second one, if I separate it into two limits, they both go towards infinity, then it's infinity - infinity?
 
Last edited:
Physics news on Phys.org
  • #2
QuarkCharmer said:

Homework Statement


[itex]lim_{x \to 0+} ln(sin(x))[/itex]

Homework Equations




The Attempt at a Solution


I'm really not sure how to take this limit at all?

I know (from using a table) that it tends towards -infinity, but I am not sure how to go about taking it manually?

I am thinking that because as x approaches 0, sin(x) approaches 0, you can treat sin(x) like x. Then as x approaches 0, ln(x) approaches -infinity.

Intuitively, that is correct. You can make it rigorous with a δ, ε argument. If you can use the fact that ln(x) → -∞ as x → 0+ then you know that given any N > 0 there is a δ > 0 such that ln(x) < -N if 0 < x < δ. Now put that together with a similar type of argument about sin(x), knowing that sin(x) → 0 as x → 0.
 
  • #3
For the first one, I think the most rigorous approach (besides epsilon-delta) is the squeeze theorem. Use

[tex]\sin(x)\leq x[/tex]

Take logs and take the limit.

For the second one, first make one logarithm using

[tex]ln(a/b)=ln(a)-ln(b)[/tex]
 

FAQ: Limit of ln(sinx) & Limit of [ln(1+x^2)-ln(1+x)]

What is the limit of ln(sinx) as x approaches 0?

The limit of ln(sinx) as x approaches 0 is 0. This can be proven using L'Hopital's rule or by graphing the function.

How do you calculate the limit of [ln(1+x^2)-ln(1+x)] as x approaches infinity?

The limit of [ln(1+x^2)-ln(1+x)] as x approaches infinity is 0. This can be shown by factoring out x^2 and then using L'Hopital's rule to evaluate the limit.

Can the limit of ln(sinx) be evaluated at x=π/2?

No, the limit of ln(sinx) cannot be evaluated at x=π/2 because sin(π/2) equals 1 and the natural logarithm of 1 is 0, which would result in an indeterminate form.

What is the significance of the limit of ln(sinx) as x approaches 0 in calculus?

The limit of ln(sinx) as x approaches 0 is an important concept in calculus because it helps to understand the behavior of functions near points of discontinuity. It also plays a role in evaluating integrals and derivatives involving trigonometric functions.

How does the limit of [ln(1+x^2)-ln(1+x)] relate to the derivative of ln(x)?

The limit of [ln(1+x^2)-ln(1+x)] is closely related to the derivative of ln(x). In fact, the limit can be interpreted as the derivative of ln(x) evaluated at x=1. This is because the derivative of ln(x) is equal to 1/x, and when x=1, this simplifies to 1. Therefore, the limit is equal to the derivative of ln(x) evaluated at x=1.

Similar threads

Replies
13
Views
1K
Replies
8
Views
1K
Replies
8
Views
1K
Replies
2
Views
3K
Replies
23
Views
2K
Back
Top